Question

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The rod AC of a TV disc antenna is fixed
at right angles to the wall AB and a rod
CD is supporting the disc .
If AC=1.5 m long and CD= 3 m,
find (i) tan theta (ii) sec theta + cosec theta.
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Answer 1

The rod AC of a TV disc antenna is fixed  at right angles to the wall AB and a rod  CD is supporting the disc .

(i) tan theta = 1/2

(ii) sec theta + cosec theta = 2

Given,

ABD is rt. angled triangle

AC (base) = 1.5 m

CD (hypotenuse) = 3 m

CD^2 = AC^2 + AD^2

3^2 = 1.5^2 + AD^2

AD^2 = 9 - 2.25 = 6.75

AD = √6.75 m

now, we need to find,

tan theta and sec theta + cosec theta

From figure, it's clear that,

tan theta = AC / DC

= 1.5 / 3

∴ tan theta = 1/2

sec theta = DC / DA = 3 / √6.75

cosec theta = DC / AC = 3 / 1.5

sec theta + cosec theta

= 3 / √6.75 + 3 / 1.5

= 3 ( 1.5 + √6.75 ) / 1.5 (√6.75)

= 9.29 / 3.89

= 2.38

∴ sec theta + cosec theta = 2.38 ≈ 2

Answer 2

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